Relativistic Electrodynamics
Continuity Equation
Recall the continuity equation ∇⋅→J=−∂ρ∂t. In terms of the four-current density, the left hand side is ∇⋅→J=∂Jx∂x+∂Jy∂y+∂Jz∂z=3∑i=1∂Ji∂xi, while the right hand side is ∂ρ∂t=1c∂J0∂t=∂J0∂x0. Thus, the continuity equation in terms of the four-current density is ∂Jμ∂xμ=0. As this is also the four-dimensional divergence of Jμ, the continuity equation implies the four-current density is divergenceless.
Maxwell's Equations in Terms of Field Tensor
Recall the Maxwell's equations ∇⋅→E=ρϵ∇×→B=μ0→J+μ0ϵ0∂→E∂t∇⋅B=0∇×→E=−∂→B∂t. For equation (1), let us examine ∂F0ν∂xν ∂F0ν∂xν=∂F00∂x0+∂F01∂x1+∂F02∂x2+∂F03∂x3=1c(∂Ex∂x0+∂Ey∂y+∂Ez∂z)=1c(∇⋅→E)=μ0cρ=μ0J0. Thus, ∂F0ν∂xν=μ0J0 is equivalent to equation (1).
For equation (2), let us examine ∂F1ν∂xν ∂F1ν∂xν=∂F10∂x0+∂F11∂x1+∂F12∂x2+∂F13∂x3=−1c2∂Ex∂t+∂Bz∂y−∂By∂z=(−1c2∂E∂t+∇×→B)x=μ0Jx=μ0J1. Similarly for the ∂F2ν∂xν and ∂F3ν∂xν. Thus, ∂Fiν∂xν=μ0Ji is equivalent to the equation (2).
For equation (3), let us examine ∂G0ν∂xν ∂G0ν∂xν=∂G00∂x0+∂G01∂x1+∂G02∂x2+∂G03∂x3=∂Bx∂x+∂By∂y+∂Bz∂z=∇⋅→B=0. Thus, ∂G0ν∂xν=0 is equivalent to equation (3).
For equation (4), let us examine ∂G1ν∂xν ∂G1ν∂xν=∂G10∂x0+∂G11∂x1+∂G12∂x2+∂G13∂x3=−1c∂Bx∂t−1c∂Ez∂y+1c∂Ey∂z=−1c(∂→B∂t+∇×→E)x=0. Similarly for the ∂G2ν∂xν and ∂G3ν∂xν. Thus, ∂Giν∂xν=0 is equivalent to equation(4).
Therefore, the Maxwell's equations in terms of the four-current density are {∂Fμν∂xν=μ0Jμ∂Gμν∂xν=0
Maxwell's Equations in terms of 4-vector Potential
Substitute Fμν=∂Aν∂xμ−∂Aμ∂xν into ∂Fμν∂xν=μ0Jμ, we have ∂∂xμ(∂Aν∂xν)−∂∂xν(∂Aμ∂xν)=μ0Jμ. As Fμν is gauge invariant (as →E and →B are gauge invariant), the gauge transformation A^\mu\to A^{\mu}'=A^\mu + \frac{\partial \lambda}{\partial x_\mu} leaves Fμν unchanged. The Lorentz gauge condiiton ∇⋅→A=−1c2∂V∂t in terms of 4-vector potential is ∂Aν∂xν=0. Therefore, (5) reduces to ◻2Aμ=−μ0Jμ, where ◻ is the d'Alembertian defined as ◻2:=∂∂xν∂∂xν=∇2−1c2∂2∂t2p
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